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Complex Numbers question

2003 · Shift 0 · Q118
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Complex Numbers question

2003 · Shift 0 · Q118

JEE MainMathematicsComplex NumbersMCQ+4 / −1
If zzz and ω\omegaω are two non-zero complex numbers such that ∣zω∣=1\left| {z\omega } \right| = 1∣zω∣=1 and Arg(z)−Arg(ω)=π2,Arg(z) - Arg(\omega ) = {\pi \over 2},Arg(z)−Arg(ω)=2π​, then  z ‾ω\,\overline {z\,} \omegazω is equal to
  1. A
    −i- i−i
  2. B
    1
  3. C
    - 1
  4. D
    iii
View written solutionFree

Correct answer: A

  1. Let z=reiα,ω=seiβz = r e^{i\alpha}, \qquad \omega = s e^{i\beta}z=reiα,ω=seiβ where r=∣z∣>0r=|z|>0r=∣z∣>0 and s=∣ω∣>0s=|\omega|>0s=∣ω∣>0.

  2. From the condition ∣zω∣=1,|z\omega|=1,∣zω∣=1, we get ∣z∣∣ω∣=rs=1.|z||\omega| = rs = 1.∣z∣∣ω∣=rs=1.

  3. Also given: Arg⁡(z)−Arg⁡(ω)=π2.\operatorname{Arg}(z)-\operatorname{Arg}(\omega)=\frac{\pi}{2}.Arg(z)−Arg(ω)=2π​. So, α−β=π2.\alpha-\beta=\frac{\pi}{2}.α−β=2π​.

  4. Now compute z‾ ω\overline{z}\,\omegazω. Since z‾=re−iα,\overline{z} = r e^{-i\alpha},z=re−iα, therefore z‾ ω=re−iα⋅seiβ=rs ei(β−α).\overline{z}\,\omega = r e^{-i\alpha} \cdot s e^{i\beta} = rs\, e^{i(\beta-\alpha)}.zω=re−iα⋅seiβ=rsei(β−α).

  5. Using rs=1rs=1rs=1 and α−β=π2\alpha-\beta=\frac{\pi}{2}α−β=2π​, β−α=−π2.\beta-\alpha = -\frac{\pi}{2}.β−α=−2π​. Hence, z‾ ω=e−iπ/2=−i.\overline{z}\,\omega = e^{-i\pi/2} = -i.zω=e−iπ/2=−i.

  6. Therefore the correct option is: −i\boxed{-i}−i​

So, Option A is correct.

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